Problem 5 - Precalculus

Question

Graph each hyperbola. Specify the eccentricity, the center, and the values of \(a, b,\) and \(c .\) (a) \(r=\frac{3}{2+4 \cos \theta}\) (b) \(r=\frac{3}{2-4 \cos \theta}\)

Step-by-Step Solution

Answer

Both hyperbolas have eccentricity \(e = 4\), and center at origin; values are \(c = 3, a = -\frac{1}{20}, b \approx 0.193\).

Step 1: Identify Hyperbola Type

The given equations are in the polar form for conic sections. We recognize the form, \(r = \frac{ed}{1 + e \cos \theta}\) or \(r = \frac{ed}{1 - e \cos \theta}\), which represents conic sections. Here, \(e\) represents the eccentricity, and \(d\) is a constant. Since the denominator includes \(\cos \theta\), this describes hyperbolas with a horizontal transverse axis.

Step 2: Analyze Equation (a)

For equation (a), \(r = \frac{3}{2 + 4 \cos \theta}\). Compare this with \(r = \frac{ed}{1 + e \cos \theta}\) to identify parameters. We have denoted numerical values: \(ed = 3\) and \(e = 4\). Therefore we solve for \(d\):\[ ed = 3 \implies 4d = 3 \implies d = \frac{3}{4} \]

Step 3: Analyze Equation (b)

For equation (b), \(r = \frac{3}{2 - 4 \cos \theta}\). Similarly, compare this with \(r = \frac{ed}{1 - e \cos \theta}\). Here, we have: \(ed = 3\) and \(e = 4\). Solving for \(d\):\[ ed = 3 \implies 4d = 3 \implies d = \frac{3}{4} \]

Step 4: Find Values of a, b, and c for Hyperbolas

For both equations, the constants can be used to solve for \(a\), \(b\), and \(c\). For hyperbolas, the relationship is given by:\[ c = ed, \quad a = \frac{d}{1-e^2}, \quad b = \frac{d}{\sqrt{e^2 - 1}} \]Substitute \(e = 4\) and \(d = \frac{3}{4}\) into these:\[ c = 4 \times \frac{3}{4} = 3 \]\[ a = \frac{\frac{3}{4}}{1 - 16} = -\frac{3}{60} = -\frac{1}{20} \]\[ b = \frac{\frac{3}{4}}{\sqrt{16 - 1}} = \frac{\frac{3}{4}}{\sqrt{15}} \approx 0.193 \]

Key Concepts

Polar Coordinates

Polar coordinates offer a different way of representing points in a plane, differing from the more familiar Cartesian coordinate system. Instead of using an \((x, y)\) pair to define a point, polar coordinates use \((r, \theta)\). Here, \(r\) is the distance from the origin to the point, and \(\theta\) is the angle formed with the positive \(x\)-axis, measured in degrees or radians.

This system is especially useful for dealing with curves that are naturally circular or radial in nature. For instance, conic sections like circles, ellipses, parabolas, and hyperbolas can often be expressed more simply in polar coordinates. These expressions often involve trigonometric functions such as \(\sin \theta\) and \(\cos \theta\).

When graphing conic sections in polar coordinates, it's important to understand the role of both \(r\) and \(\theta\). Adjusting \(r\), which is always non-negative, changes how far a point is from the origin. Changing \(\theta\) rotates the point around the origin.

Conics in Polar Form

Conics in polar form are a powerful way to describe the family of curves known as conic sections, which include circles, ellipses, parabolas, and hyperbolas. The general polar equation for a conic section is given by \( r = \frac{ed}{1 + e \cos \theta} \) or \( r = \frac{ed}{1 - e \cos \theta} \).

These equations use the concept of eccentricity, represented by \(e\), to define the shape of the conic. Each conic type has a characteristic range for \(e\):

• A circle has \(e=0\).
• An ellipse has \(0
• A parabola has \(e=1\).
• A hyperbola has \(e>1\).

For hyperbolas, which are our focus here, the formula alters based on whether we add or subtract \(e \cos \theta\) in the denominator. Interestingly, switching from addition to subtraction alters the orientation of the conic, thus changing which axis—horizontal or vertical—it is centered on. This modification is crucial for correctly interpreting the graph's layout in polar coordinates.

Eccentricity

Eccentricity, denoted as \(e\), is a crucial parameter in polar equations of conic sections, giving insight into the shape and peculiarities of the curve. It determines the "flatness" or "openness" of the conic section:

• When \(e = 0\), the conic is a perfect circle, perfectly symmetrical around the center.
• For \(0 < e < 1\), the conic is an ellipse, still closed but stretched more in one direction.
• If \(e = 1\), we have a parabola, which is open-ended.
• For \(e > 1\), the conic is a hyperbola, characterized by two separate curves.

In our problem, the eccentricity \(e = 4\) indicates it's a hyperbola. Eccentricity dictates how the \(r\) value will change as \(\theta\) changes, directly affecting the curve's openness and direction. The larger the \(e\), the more "open" the hyperbola is, meaning its branches will diverge more steeply from the center.

Transverse Axis

The transverse axis is a key geometric feature when analyzing hyperbolas. It refers to the line segment that passes through the center of the hyperbola and connects its two vertices. This axis is crucial because it helps determine the orientation and position of the hyperbola.

For hyperbolas expressed in polar form, particularly those of the form \( r = \frac{ed}{1 \pm e \cos \theta} \), the transverse axis is usually horizontal, especially when the equation uses \(\cos \theta\) as opposed to \(\sin \theta\). The form of the equation highlights whether the transverse axis aligns horizontally or vertically.

Understanding the transverse axis gives visual clarity about the "direction" the hyperbola opens. In our exercise, the polar equations suggest a transverse axis that aligns horizontally due to the inclusion of \(\cos \theta\), effectively influencing how we interpret the spatial configuration of the hyperbola's branches.